Definitions and Calculations — Page 58, Lesson 76

All right, let's get into the heart of mass and balance work. This is where the theory turns into the actual numbers you'll be moving around on a load sheet. We're looking at a set of problems about adding or removing mass, and the core idea behind all of them is a single, powerful relationship. Here's the principle: the centre of gravity, the CG, is the balance point of the aircraft. When you add or remove mass at some distance from the datum, you're applying a moment—a turning force—that shifts that balance point. The key formula that governs all of this is that the change in CG position is equal to the moment change divided by the final total mass. Let me break that down. The moment change is simply the mass you're adding or removing, multiplied by its distance from the datum. And the final total mass is the aircraft's original mass, plus the mass you add, or minus the mass you remove. So, if you're removing mass, the moment change is negative, and the final mass is the original minus the removed mass. If you're adding, both are positive. Now, let's walk through the problems to see how this works in practice. I want you to watch how we set up the signs and the distances. Problem 1: We have three holds, A, B, and C, at 10, 100, and 250 inches aft of the datum. The aircraft's total mass is 3500 kg, and the CG is at 70 inches aft. The CG limits are from 40 to 70 inches aft. We need to remove load from hold C to bring the CG onto the forward limit, which is 40 inches. So, the CG must move forward, from 70 inches to 40 inches—a change of 30 inches forward. We're removing mass from hold C, which is at 250 inches aft. So, the moment change is the removed mass, let's call it m, times 250 inches. The final mass is 3500 minus m. Setting up the formula: the change in CG, which is 30 inches, equals m times 250, divided by (3500 minus m). Solving that gives us the mass to remove. Problem 2: Here, the mass is 5000 lb, and the CG is at 80 inches aft. The aft limit is 80.5 inches. We're adding mass to a hold at 150 inches aft. The CG can only move aft by 0.5 inches before hitting the limit. So, the change in CG is 0.5 inches. The moment change is the added mass, m, times 150 inches. The final mass is 5000 plus m. So, 0.5 equals m times 150, divided by (5000 plus m). Solve for m, and that's the maximum mass you can add. Problem 3: Now we have a loaded mass of 108,560 lb, and the CG is at 86.3 feet aft. The aft limit is 85.6 feet. We need to place ballast in a hold at 42 feet aft to bring the CG forward onto the aft limit. Wait, let me check that. The CG is at 86.3, and the aft limit is 85.6. So, the CG is aft of the limit. We need to move it forward, to 85.6 feet. That's a change of 0.7 feet forward. We're adding ballast at 42 feet aft. The moment change is the ballast mass, m, times 42 feet. The final mass is 108,560 plus m. So, 0.7 equals m times 42, divided by (108,560 plus m). Solve for m. Problem 4: This is similar to problem 2, but we're removing mass. The aft limit is 80 inches, and the loaded CG is at 80.5 inches—again, aft of the limit. The mass is 6400 lb. We remove mass from a hold at 150 inches aft. We need to move the CG forward by 0.5 inches. The moment change is the removed mass, m, times 150 inches, but it's negative because we're removing it. The final mass is 6400 minus m. So, 0.5 equals m times 150, divided by (6400 minus m). Solve for m. Problem 5: This one asks for the new CG position after adding a package. The mass is 7900 kg, and the CG is at 81.2 inches aft. We add a 250 kg package at 32 inches aft. The moment change is 250 times 32. The final mass is 7900 plus 250, which is 8150 kg. The change in CG is 250 times 32, divided by 8150. That gives you the shift. Since the package is at 32 inches, which is forward of the current CG at 81.2, the CG will move forward. So, you subtract that change from 81.2 to get the new CG position. Problem 6: Here, the CG limits are from 72 to 77 inches aft. The mass is 3700 kg, and the CG is at 76.5 inches. We remove 60 kg from the forward hold, which is at 147 inches forward of the datum. Note that—it's forward, so the distance is negative in our moment calculation. The moment change is minus 60 kg times minus 147 inches, which is a positive moment change. The final mass is 3700 minus 60, which is 3640 kg. The change in CG is that moment change divided by 3640. Since the moment change is positive, the CG moves aft. So, you add that change to 76.5 to find the new position. Problem 7: This one is cut off, but it starts with a zero fuel mass of 47,800 kg and a performance limited take-off—likely a take-off mass limit. The same principle will apply: you'll be adding fuel or payload, and you'll need to calculate the CG shift using the same formula. Now, one thing I want to stress: in every one of these, the distance is measured from the datum, and the sign—forward or aft—is critical. Aft of the datum is positive, forward is negative. And when you remove mass, the moment is negative; when you add, it's positive. Get those signs right, and the formula does the rest. Let's also look at the figures that accompany this. shows the concept of adding or removing mass visually. And shows a specific problem about transferring load between holds—that's a slightly different scenario, where you're moving mass from one hold to another, which changes the moment without changing the total mass. So, the takeaway is this: the change in CG equals the moment change divided by the final mass. Master that, and you can solve any of these. Let's work through the numbers together if you want to check your arithmetic.